It is given in the question that,

Height of the Cylindrical portion (H) $=13cm$

Radius of the Cylindrical portion (r) $=5cm$

Height of the whole solid $=30cm$

Now,

The curved surface area of the cylinder $\left( S_{1}^{{}} \right)=2\pi rh$

${{S}_{1}}=2\pi \left( 5 \right)\left( 13 \right)$

${{S}_{1}}=408.2c{{m}^{2}}$

Assume, ‘L’ be the slant height of the cone

Now, the curved surface area of the cone $\left( {{S}_{2}} \right)=\pi rL$

${{S}_{2}}=\pi \left( 6 \right)L$

For conical part, it is given that,

$h=30–13–5=12cm$

So, we all know that

${{L}^{2}}={{r}^{2}}+{{h}^{2}}$

${{L}^{2}}={{5}^{2}}+{{12}^{2}}$

${{L}^{2}}=25+144$

${{L}^{2}}=169$

$L=13m$

Now,

${{S}_{2}}=\pi \left( 5 \right)\left( 13 \right)c{{m}^{2}}$

${{S}_{2}}=204.28c{{m}^{2}}$

Then, the curved surface area of the hemisphere $\left( {{S}_{3}} \right)=2\pi {{r}^{2}}$

${{S}_{3}}=2\pi {{\left( 5 \right)}^{2}}$

${{S}_{3}}=157.14c{{m}^{2}}$

Thus, the total curved surface area $\left( S \right)={{S}_{1}}+{{S}_{2}}+{{S}_{3}}$

$S=(408.2+204.28+157.14)$

$S=769.62c{{m}^{2}}$

Hence, the surface area of the toy is $770c{{m}^{2}}$